The equation of the straight line whose slope | vedclass

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Question

(A) Let the points be $A(1, 2)$ and $B(-3, 5)$. The point $P(x, y)$ divides the segment $AB$ externally in the ratio $m:n = 4:3$.Using the section for

Vedclass · Accepted Answer

$2x + 3y - 12 = 0$

Vedclass · Answer

(A) Let the points be $A(1, 2)$ and $B(-3, 5)$. The point $P(x, y)$ divides the segment $AB$ externally in the ratio $m:n = 4:3$.Using the section formula for external division: $x = \frac{mx_2 - nx_1}{m - n}$ and $y = \frac{my_2 - ny_1}{m - n}$.$x = \frac{4(-3) - 3(1)}{4 - 3} = \frac{-12 - 3}{1} = -15$.$y = \frac{4(5) - 3(2)}{4 - 3} = \frac{20 - 6}{1} = 14$.So,the point is $(-15, 14)$.The equation of the line with slope $m = \frac{-2}{3}$ passing through $(-15, 14)$ is given by $y - y_1 = m(x - x_1)$.$y - 14 = \frac{-2}{3}(x + 15)$.$3y - 42 = -2x - 30$.$2x + 3y - 12 = 0$.

The equation of the straight line whose slope is $\frac{-2}{3}$ and which divides the line segment joining $(1, 2)$ and $(-3, 5)$ in the ratio $4:3$ externally is

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